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ch06_ism - 6 Inductance Capacitance and Mutual Inductance...

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6 Inductance, Capacitance, and Mutual Inductance Assessment Problems AP 6.1 [a] i g = 8 e 300 t 8 e 1200 t A v = L di g dt = 9 . 6 e 300 t + 38 . 4 e 1200 t V , t > 0 + v (0 + ) = 9 . 6 + 38 . 4 = 28 . 8 V [b] v = 0 when 38 . 4 e 1200 t = 9 . 6 e 300 t or t = (ln 4) / 900 = 1 . 54 ms [c] p = vi = 384 e 1500 t 76 . 8 e 600 t 307 . 2 e 2400 t W [d] dp dt = 0 when e 1800 t 12 . 5 e 900 t + 16 = 0 Let x = e 900 t and solve the quadratic x 2 12 . 5 x + 16 = 0 x = 1 . 45 , t = ln 1 . 45 900 = 411 . 05 µ s x = 11 . 05 , t = ln 11 . 05 900 = 2 . 67 ms p is maximum at t = 411 . 05 µ s [e] p max = 384 e 1 . 5(0 . 41105) 76 . 8 e 0 . 6(0 . 41105) 307 . 2 e 2 . 4(0 . 41105) = 32 . 72 W [f] i max = 8[ e 0 . 3(1 . 54) e 1 . 2(1 . 54) ] = 3 . 78 A w max = (1 / 2)(4 × 10 3 )(3 . 78) 2 = 28 . 6 mJ [g] W is max when i is max, i is max when di/dt is zero. When di/dt = 0 , v = 0 , therefore t = 1 . 54 ms. 6–1
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6–2 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance AP 6.2 [a] i = C dv dt = 24 × 10 6 d dt [ e 15 , 000 t sin 30 , 000 t ] = [0 . 72 cos 30 , 000 t 0 . 36 sin 30 , 000 t ] e 15 , 000 t A , i (0 + ) = 0 . 72 A [b] i π 80 ms = 31 . 66 mA , v π 80 ms = 20 . 505 V , p = vi = 649 . 23 mW [c] w = 1 2 Cv 2 = 126 . 13 µ J AP 6.3 [a] v = 1 C t 0 i dx + v (0 ) = 1 0 . 6 × 10 6 t 0 3 cos 50 , 000 x dx = 100 sin 50 , 000 t V [b] p ( t ) = vi = [300 cos 50 , 000 t ] sin 50 , 000 t = 150 sin 100 , 000 t W , p (max) = 150 W [c] w (max) = 1 2 Cv 2 max = 0 . 30(100) 2 = 3000 µ J = 3 mJ AP 6.4 [a] L eq = 60(240) 300 = 48 mH [b] i (0 + ) = 3 + 5 = 2 A [c] i = 125 6 t 0 + ( 0 . 03 e 5 x ) dx 2 = 0 . 125 e 5 t 2 . 125 A [d] i 1 = 50 3 t 0 + ( 0 . 03 e 5 x ) dx + 3 = 0 . 1 e 5 t + 2 . 9 A i 2 = 25 6 t 0 + ( 0 . 03 e 5 x ) dx 5 = 0 . 025 e 5 t 5 . 025 A i 1 + i 2 = i AP 6.5 v 1 = 0 . 5 × 10 6 t 0 + 240 × 10 6 e 10 x dx 10 = 12 e 10 t + 2 V v 2 = 0 . 125 × 10 6 t 0 + 240 × 10 6 e 10 x dx 5 = 3 e 10 t 2 V v 1 ( ) = 2 V , v 2 ( ) = 2 V W = 1 2 (2)(4) + 1 2 (8)(4) × 10 6 = 20 µ J
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Problems 6–3 AP 6.6 [a] Summing the voltages around mesh 1 yields 4 di 1 dt + 8 d ( i 2 + i g ) dt + 20( i 1 i 2 ) + 5( i 1 + i g ) = 0 or 4 di 1 dt + 25 i 1 + 8 di 2 dt 20 i 2 = 5 i g + 8 di g dt Summing the voltages around mesh 2 yields 16 d ( i 2 + i g ) dt + 8 di 1 dt + 20( i 2 i 1 ) + 780 i 2 = 0 or 8 di 1 dt 20 i 1 + 16 di 2 dt + 800 i 2 = 16 di g dt [b] From the solutions given in part (b) i 1 (0) = 0 . 4 11 . 6 + 12 = 0; i 2 (0) = 0 . 01 0 . 99 + 1 = 0 These values agree with zero initial energy in the circuit. At infinity, i 1 ( ) = 0 . 4 A ; i 2 ( ) = 0 . 01 A When t = the circuit reduces to · . . i 1 ( ) = 7 . 8 20 + 7 . 8 780 = 0 . 4 A ; i 2 ( ) = 7 . 8 780 = 0 . 01 A From the solutions for i 1 and i 2 we have di 1 dt = 46 . 40 e 4 t 60 e 5 t di 2 dt = 3 . 96 e 4 t 5 e 5 t Also, di g dt = 7 . 84 e 4 t Thus 4 di 1 dt = 185 . 60 e 4 t 240 e 5 t
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6–4 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 25 i 1 = 10 290 e 4 t + 300 e 5 t 8 di 2 dt = 31 . 68 e 4 t 40 e 5 t 20 i 2 = 0 . 20 19 . 80 e 4 t + 20 e 5 t 5 i g = 9 . 8 9 . 8 e 4 t 8 di g dt = 62 . 72 e 4 t Test: 185 . 60 e 4 t 240 e 5 t 10 290 e 4 t + 300 e 5 t + 31 . 68 e 4 t 40 e 5 t +0 . 20 + 19 . 80 e 4 t 20 e 5 t ?
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